Question

Q3) The sides of a quadrilateral field, taken in order are 26 m, 27 m, 7 m, 24 m respectively. The angle contained by the last two sides is a right angle. Find its area​

Answer 1

Step-by-step explanation:

The sides of a quadrilateral field taken order as

AB=26 m

BC=27 m

CD=7 m

and DA=24 m

Diagonal AC is joined

Now △ADC

AC

2

=AD

2

+CD

2

AC=

24

2

+7

2

AC=

625

AC=25 m

Now area of △ABC

S=

2

1

(AB+BC+CA)

=

2

1

(26+27+25)

=

2

78

=39 m

By using heron's formula

Area of △ABC=

S(S−AB)(S−BC)(S−CA)

=

39(39−26)(39−27)(39−25)

=

39×13×12×14

=

85179

=291.85 m

2

Now area of △ADC

S=

2

1

(AD+CD+AC)

=

2

1

(25+24+7)

=

2

56

=28 m

By using heron's formula

Area of △ADC=

S(S−AD)(S−DC)(S−CA)

=

28(28−24)(28−7)(28−25)

=

28×4×21×3

=

7056

=84 m

2

hence, total area is375.85 m

2

solution